$\left| {\begin{array}{ccc} 1/a & a^2 & bc \\ 1/b & b^2 & ca \\ 1/c & c^2 & ab \end{array}} \right| = $

  • A
    $abc$
  • B
    $1/abc$
  • C
    $ab + bc + ca$
  • D
    $0$

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Similar Questions

નિશ્ચાયકના ગુણધર્મોનો ઉપયોગ કરીને અને વિસ્તરણ કર્યા વગર સાબિત કરો કે:
$\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right|=0$

$\left| {\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 4}\\{x + 3}&{x + 5}&{x + 8}\\{x + 7}&{x + 10}&{x + 14}\end{array}} \right| = $

જો ${D_r} = \left| \begin{array}{ccc} {2^{r - 1}} & {2 \cdot 3^{r - 1}} & {4 \cdot 5^{r - 1}} \\ x & y & z \\ {2^n} - 1 & {3^n} - 1 & {5^n} - 1 \end{array} \right|$ હોય,તો $\sum\limits_{r = 1}^n {D_r} = $ ની કિંમત શોધો.

Difficult
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નિશ્ચાયકનું મૂલ્ય શોધો: $\left| \begin{array}{ccc} 1/a & 1 & bc \\ 1/b & 1 & ca \\ 1/c & 1 & ab \end{array} \right|$

જો $\left| \begin{array}{ccc} a^2 & b^2 & c^2 \\ (a + \lambda)^2 & (b + \lambda)^2 & (c + \lambda)^2 \\ (a - \lambda)^2 & (b - \lambda)^2 & (c - \lambda)^2 \end{array} \right| = k\lambda \left| \begin{array}{ccc} a^2 & b^2 & c^2 \\ a & b & c \\ 1 & 1 & 1 \end{array} \right|, \lambda \neq 0$ હોય,તો $k$ ની કિંમત શોધો.

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